Question

在这段代码中，我有一个包含来自'a''b''c''d'的大量组合的数据，我试图找出每个组合存在的频率。（数据示例：abdc，abcc ，abcd，abbb，aaaa，abdc，...）之后我希望得到每个字母组合的0％到100％的答案。如果它是零。

示例输入：

letters: ['abc','aaa','abb','acc','aac','abc','bbb','ccc','ddd','abc','adc','acd','acd','aac','aad','bba','bab','abb','abc','abd'...]

我从这里得到df :( tab_files是获取我的数据的文件）

for i, tab_file in enumerate(tab_files):
    df = pd.DataFrame.from_csv(tab_file, sep='\t')

这是我的尝试：

#letter_l = all combinations of letters (abcd) together
nt_l = "abcd"
letter_l = []
for i1 in nt_l:
    for i2 in nt_l:
        for i3 in nt_l:
            letter = i1+i2+i3
            letter_l.append(letter)
#print(letter_l)

#calculates the amount of each letter combination and shows the percentage
x = []
number_per_combination = {}
for b in letter_l:    
    counter = 0
    number_per_combination[b] = 0
    for c2 in df.letter:
        if c2 == b:
           counter +=1
           number_per_combination[b] += 1
 # amount of each letter combination divided through the whole amount
    x.append(counter/(len(df.letter)))

但我得到了奇怪的百分比作为答案......我不明白为什么。有人可以帮帮我吗？

Output I want:     number_per combination
'abc': 20%        (40)
'aaa': 10%        (20)
'ccd': 0%         (0)
'ddd': 3%         (6)...

Answer 1

那么你要做的是直方图？这是一个简单的方法：

input_list = ['a', 'a', 'b', 'b', 'b', 'c']

def histogram(my_list):
    result = {}
    for item in my_list:
        result[item] = result.get(item, 0) + 1
    return result

print(str(histogram(input_list)))

.get（）方法从字典中返回给定键的值。如果密钥不存在，则插入并给出第二个参数中提供的值。

Answer 2

import re
import itertools

data="aaa, abc, aab"
words = re.split(', ',data)
words_count = {}
total_count = len( words )

for word in list(itertools.product(["a","b","c","d"], repeat=3)):
  words_count["".join(word)] = 0

for word in words:
  words_count[word] = words_count.get(word,0) + 1

for word in words_count:
  p = words_count[word]/total_count * 100
  print( "%s: %.3f%%\t(%d)" % (word,p,words_count[word]) )

计算Python中的百分比（0％到100％）

2 个答案: